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Notes on1.63 Advanced Environmental Fluid Mechanics

Instructor: C. C. Mei, 2002ccmei@mit.edu, 1 617 253 2994

October 21, 20024-1-2energy.tex

Chapter 4. Thermal effects on fluid flow

4.1 Heat and energy conservation

Recall the basic equations for a compressible fluid. Mass conservation requires that :

ρt +∇ · ρ~q = 0 (4.1.1)

Momentum conservation requires that :

ρ (~qt + ~q∇ · ~q) = −∇p+∇·=τ +ρ~f (4.1.2)

where the viscous stress tensor=τ has the components

³=τ´ij= τij = µ

Ã∂qi∂xj

+∂qi∂xi

!+ λ

∂qk∂xk

δij

There are 5 unknowns ρ, p, qi but only 4 equations. One more equation is needed.

4.1.1 Conservation of total energy

Consider both mechanical ad thermal energy. Let e be the internal (thermal) energy perunit mass due to microscopic motion, and q2/2 be the kinetic energy per unit mass due tomacroscopic motion. Conservation of energy requires

D

Dt

ZZZVρ

Ãe+

q2

2

!dV rate of incr. of energy in V (t)

= −ZZS

~Q · ~n dv rate of heat flux into V

+ZZZ

Vρ~f · ~q dV rate of work by body force

+ZZX

~Σ · ~q dS rate of work by surface force

2

Use the kinematic transport theorm, the left hand side becomes

ZZZVρD

Dt

Ãe+

q2

2

!dV

Using Gauss theorem the heat flux term becomes

−ZZSQinidS = −

ZZZV

∂Qi∂xi

dV

The work done by surface stress becomesZZSΣjqj dS =

ZZS(σjini)qj dS

=ZZS(σijqj)ni dS =

ZZZV

∂(σijqj)

∂xidV

Now all terms are expressed as volume integrals over an arbitrary material volume, thefollowing must be true at every point in space,

ρD

Dt

Ãe+

q2

2

!= −

∂Qi∂xi

+ ρfiqi +∂(σijqi)

∂xj(4.1.3)

As an alternative form, we differentiate the kinetic energy and get

ρDe

Dt+ ρqi

Dq1Dt| {z } = ρfiqi + qi

∂σij∂xj| {z }−

∂Qi∂xi

− p∂qi∂xi

+∂(τijqi)

∂xj(4.1.4)

Becasue of momentum conservation, the terms inhluded in the underbraces cancel, leaving

ρDe

Dt= −

∂Qi∂xi

− p∂qi∂xi

+∂(τijqi)

∂xj(4.1.5)

We must now add Fick’s law of heat conduction

Qi = −K∂T

∂xi(4.1.6)

where K is the heat conductivity, and the following equations of state

e = e(p, T ) (4.1.7)

ρ = ρ(p, T ) (4.1.8)

Now there are 10 unknowns qi, Qi, ρ, p, e, T , and 10 equations: 1 from (4.1.1), 3 from (4.1.2),1 from (4.1.5), 3 from (4.1.6), 1 from (4.1.7) and 1 from (4.1.8).

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4.1.2 Equations of state:

For a perfect gas:

e = CvT (4.1.9)

p = ρRT, where R = Cp − Cv. (4.1.10)

The specific heats Cv (constant volume) and Cp (constatn pressure) are measured in Joules/kg -dyne.For a liquid:

e = CT (4.1.11)

ρ = ρo³To, po

´+

∂ρ

∂p

¯¯T

∆p+∂ρ

∂T

¯¯p

∆T + · · ·

where ρo, To, po are some constant reference density, temperature, and pressure respectively,while ∆p = p− po and ∆T = T − To. are the variations in pressure and temperature. Thesevariations are usually small in environmental problems. We define the thermal expansioincoefficient β by :

β = −1

ρ

∂ρ

∂T

¯¯p

=1

V

∂V

∂T

¯¯p

(4.1.12)

where V denotes the specific volume (volume per unit mass), and the bulk modulus 1/ε by

ε = −1

ρ

∂ρ

∂p

¯¯T

= −1

V

∂V

∂p

¯¯T

(4.1.13)

ρ ∼= ρo(1− β∆T + ε∆p) (4.1.14)

For liquids

O(β) ∼ 10−3/oK, O(ε) ∼ 10−6/atm

Rewrite Eqn. (4.1.5)

ρDe

Dt+ ρqi

DqiDt

= ρqifi + qi∂σij∂xj

+∂

∂xiK∂T

∂xi− p

∂qi∂xi

+ τij∂qi∂xj

(4.1.15)

In summary, we have, for a perfect gas

ρCvDT

Dt= −p

∂qi∂xi

+∂

∂xiK∂T

∂xi+ τij

∂qi∂xj

(4.1.16)

and for an incompressible liquid

ρCDT

Dt=

∂

∂xiK∂T

∂xi+ τij

∂qi∂xj

(4.1.17)

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Recall that

Φ = τij∂qi∂xj

(4.1.18)

is the rate of viscous dissipation. Thus the rate of change in internal energy is equal to thesum of rate of pressure working to compress the fluid, viscous dissipation, and heat diffusion.More will be said about the incompressiblilty of liquid later.

4.2 Approximations for small temperature variation

4.2.1 Mass conservation and almost incompressibilty :

Recall the law of mass conservation:

−1

ρ

Dρ

Dt= ∇ · ~q

Let the time scale be L/U . The left-hand-side is of the order UL∆ρρwhile the right-hand-side

is UL. For ∆T = O(10oC), their ratio is

∆ρ

ρ∼∆T

T∼10oK

300oK¿ 1

Therefore,∇ · ~q = 0. (4.2.1)

The fluid is approximately incompressible even if ∆T 6= 0.

4.2.2 Momentum conservation and Boussinesq approximation

In static equilibirum ~qo ≡ 0. Therefore,

−∇po + ~fρo = 0. (4.2.2)

Let p = pd + po where pd is the dynamic pressure

ρ = ρd + ρo

−∇p+ ρ~f = −∇po + ρo ~f −∇pd + (ρ− ρo) ~f

Therefore,

ρD~q

Dt= −∇pd +∇·

=τ + (ρ− ρo) ~f| {z }

buoyancy force

(4.2.3)

Nowρ = ρo[1− β(∆To +∆Td)]) (4.2.4)

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Henceρo = ρo(1− β∆To), ρd = −ρoβ∆Td,

and(ρ− ρo)~f = −ρo(−g)β∆Td~k = ρogβ∆Td~k (4.2.5)

For mildly varying ρo and small ρ− ρo, we ignore the variation of density and approximateρo by a constant everywhere, except in the body force. This is called the Boussinesqapproximation. Thus

ρoD~q

Dt= −∇pd +∇·

=τ +ρogβ∆Td ~k (4.2.6)

whereρo = ρo(z = 0)

4.2.3 Total energy

Using Eqn. (4.2.1) in Eqn. (4.1.17) and the Boussinesq approximation

ρoCDT

Dt=

∂

∂xiK∂T

∂xi+ Φ (4.2.7)

Here T is the total temperature (static + dynamic).Now

Φ

ρoCDTDt

∼µU2/L2

ρoCU∆TL

∼µ

ρoUL

U2

C∆T=E

Re

where

E =U2

C∆T= Eckart No., Re =

ρUL

µ= Reynolds No.

In environmental problems, ∆T ∼ 10 oK,L ∼ 10 m,U ∼ 1m/sec, the last two columns of

Table 4.1: Typical values E/Re for air and water

Water AirC (erg/s-gr-◦K) 4 × 107 107

K (ergs-cm-◦K) 0.6 × 105 0.3 × 105

ν(cm2/s) 102 2 × 10−2

β(1/◦K) 10−3 1/300E 0.25 × 10−2 10−4

Re 105 0.5 × 105

Table 4.1 is obtained. Hence Φ is negligible, and

ρoCDT

Dt=

∂

∂xiK∂T

∂xi(4.2.8)

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Only convection and diffusion are dominant. This is typical in natural convectin problems.Remark 1. In many engineering problems (aerodynamics, rocket reentry, etc.), heat

is caused by frictional dissipation in the flow, therefore, Φ is important. These are calledforced convection problems. In environmental problems, flow is often the result of heataddition. Here we have thenatural convection.Remark 2: Since T appears as a derivative only, only the variation of T , i.e., the

difference T − To matters, where To is a reference temperature.Remark 3: In turbulent natural convection

u = u+ u0 T = T + T 0 (4.2.9)

Averaging Eqn. (4.2.8)

ρocDT

Dt= − ρoc

∂

∂xiu0iT 0| {z }

heat flux by turbulence

+∂

∂xiK∂T

∂xi(4.2.10)

If the the correlation term is modeled as eddy diffusion, the form would be similar to (4.2.8).